1

CE 366 – SHALLOW FOUNDATIONS CANTILEVER FOOTING Given: Q1 = 1500 kN, Q2 = 2500 kN, qall = 200 kN/m2

Design a cantilever footing for a uniform soil pressure distribution.

Locate ΣQ = Q1 + Q2

b=1500 × 8 / 4000=3

Select B1=2m

a = 8.0 + 0.4 – 1.0 – 1.0

a = 4.40m

R1= (4000 × 3) / 7.4 = 1620kN

R2 = 4000 – 1620 = 2380 kN

Determine B2,

qall = Q / (2 × B2)

200 = 1620 / (2 × B2) B2 = 4 m

Similary,

200 = 2380 / B2 B = 3.45 m

L=8 m

Q1 = 1500 kN Q2 = 2500 kN

B1=2m

B

B

B2 Strap

0.4m

Q1+Q2 =4000kN

1.0

a=4.40 b=3.0

a+b=7.4

200 kN/m2 200 kN/m2

R2=2380 kN R1=1620

Q1=1500 Q2-R2=120 kN

R1=1620

SHEAR

MOMENT

1500kN 120kN

1500x0.6 =900 kN-m

Q1+Q2 = R1+R2

0.8 m 0.8 m

0.6m

2

TRAPEZOIDAL FOOTING Consider weight of footing and determine B1 and B2 of a trapezoidal footing for a uniform soil

pressure of 300 kN/m2. (γconc = 24kN/m3)

Centroid of the base

Weight of footing= 16x24x(B1+B2)/2 = 192(B1+B2)

Area of footing = 8(B1+B2)

ΣQ=0 9000+192(B1+B2)=8(B1+B2)x300

B1+B2 = 4.07m ………(1)

ΣM=0 (moment about centroid of the base)

3000(14-x)+600+2000(8-x)+800-4000(x-2)-1200+(wght of ftg)x0=(base pressure)x0

x = 7.36m

21

21

BBBB2x16x

3136.7x

++

== B2= 1.63B1 ………. (2)

From (1) and (2) B1= 1.55m ; B2 = 2.52m

2 6 6 2

1

M1=600kN-m Q1=3000kN

M2=800kN-m Q2=2000kN

M3=1200kN-m Q3=4000kN

B1 B2

x

ΣQ=9000kN

3

MAT FOUNDATION A mat foundation rests on a sand deposit whose allowable bearing value is 150 kN/m2. Column

loads are given in the figure. The thickness of the mat is 2.0 m (γconcrete = 24 kN/m3). Calculate

base pressures assuming that the lines passing through the centroid of the base and parallel to the

sides are principal axes. Is the mat foundation given safe?

L = 28 m

6 m 6 m 6 m 6 m

6 m

6 m

2 m

2 m 2 m

2 m

1000 kN 2000 kN 3000 kN 2000 kN 1000 kN

2500 kN 3500 kN 4000 kN 3500 kN 2500 kN

1000 kN 2000 kN 3000 kN

B =

16 m

Ä = 12 m

u =

6 m

EXISTING STRUCTURE

4

Area of foundation = 28 × 16 – 12 × 6 = 376 m2

Total vertical load = Σ V = Column loads + (376) × 24 × 2 = 49048 kN * Center of gravity (CG) of base :

* Location of ΣQ :

Take moment about the left side:

= (1 / 49048) . [ 2 × (1000+2500+1000) + 8 × (2000+3500+2000) + 14 × (3000 +

+ 4000 + 3000) + 20 × (3500+2000) + 26 × (2500+1000) + 376 × 2 × 24 × 12.47 ]

= 12.95 m from left

Take moment about bottom side :

= (1 / 49048) . [ 2 × (1000+2000+3000+2000+1000) + 8 × (2500 + 3500 + 4000 +

+ 3500 + 2500) + 14 × (1000+2000+3000) + 376 × 2 × 24 × 7.04 ]

= 7.28 m from bottom

47.12376

)616(126141628=

+××−×× m from left

04.7376

)103(12681628=

+××−×× m from bottom

10

3

16 6

•

•

•12.47

7.04

CG

5

Eccentricity :

e1 = 12.95 – 12.47 = 0.48 m e2 = 7.28 – 7.04 = 0.24 m

M1 about 1-1 axis: M1 = ΣQ ⋅ e1 = 49048 ⋅ (0.48) = 23543 kN.m

M2 about 2-2 axis:

M2 = ΣQ ⋅ e2 = 49048 ⋅ (0.24) = 11772 kN.m

•

12.47

7.04

1

1

2 2

15.53

8.96

28 m

M1

16 m

M2

=⎥⎦

⎤⎢⎣

⎡⋅⋅+

⋅−⎥

⎦

⎤⎢⎣

⎡⋅⋅+

⋅=−

21

32

1

3

11 )(12

)(12

dlblbDLBLBI

22915)47.1222(12612126)47.1214(2816

122816 2

32

3

=⎥⎦

⎤⎢⎣

⎡−××+

×−⎥

⎦

⎤⎢⎣

⎡−××+

×= m4

=⎥⎦

⎤⎢⎣

⎡⋅⋅+

⋅−⎥

⎦

⎤⎢⎣

⎡⋅⋅+

⋅=−

22

32

2

3

22 )(12

)(12

dlbblDLBBLI

7197)04.713(61212

612)04.78(281612

1628 23

23

=⎥⎦

⎤⎢⎣

⎡−××+

×−⎥

⎦

⎤⎢⎣

⎡−××+

×= m4

6

Note: In soil mechanics compression is taken as positive (+)

qE = 134.9 kN/m2 qF = 151.3 kN/m2

Since at all critical points stress values are almost < qall = 150 kN/m2 given mat foundation is safe.

2 2

1

1

CG

ΣQ = 49048 kN

ED

C B

AF e2 = 0.24 m

e1 = 0.48 m

qF = 151.3 kN/m2

qE = 134.9 kN/m2

qD = 106 kN/m2

qC = 132.2kN/m2 qB = 148.7 kN/m2

qA = 138.9 kN/m2

(MAX)

(MIN)

22

22

11

11

−−

⋅±

⋅±

Σ=

IyM

IyM

AreaQq

2121 64.103.14.130

719711772

2291523543

37649048 yyyyq ±±=

⋅±

⋅±=

9.138)96.2(64.1)53.3(03.14.13064.103.14.130 21 =⋅+⋅+=±±= yyqAkN/m2

7.148)96.8(64.1)53.3(03.14.130 =⋅+⋅+=Bq kN/m2

2.132)96.8(64.1)47.12(03.14.130 =⋅+⋅−=Cq kN/m2

106)04.7(64.1)47.12(03.14.130 =⋅−⋅−=Dq kN/m2

7

COMBINED FOOTING ANALYZED BY RIGID METHOD A rectangular combined footing which supports three columns is to be constructed on a sandy clay layer whose

allowable bearing value (base pressure) is 84 kN/m2( qall is given in terms of net stresses). The thickness of the

concrete footing is 0.85m. There is a 1.20m thick soil fill on the footing. Unit weight of the soil and the concrete are

20 kN/m3 ans 24 kN/m3 respectively. Analyze the footing by rigid method and plot shear and moment diagrams. (γsoil

= 20 kN/m3; γconc = 24 kN/m3)

Neglecting column weights;

ΣQnet=800+600+500+(24-20)x16xBx0.85 = 1900+54.4B

BBxx

VMe

4.5419001800

4.54190065006800

+=

+−

=ΣΣ

=

mBxB

xBx

B

mkNLe

BxLVq

0.2)16)4.541900(

180061(16

4.54190084

/84)61( 2max

=⇒+

++

=

=+Σ

=

Use B=2.0m qmax = 84 kN/m2 ; qmin = 41.6 kN/m2

Downward uniform pressure = 0.85 × (24-20) = 3.4 kN/m2

800kN 600kN 500kN

2m 6m 2m 6m B=?

1.2

0.85

8

These are the diagrams related to forces and moments acting on the foundation. Explanations are

at the next page;

800kN

41.6 - 3.4 = 38.2

(shear diagram is parabolic in fact, but we linearized it for simplicity)

600kN 500kN

2m 6m 2m 6m

84 41.6

3.4 3.4

59.4 43.5 75.2

4.2

7.7

1.01

h

g

f

e

d c

b a

311.4 319.2 336

280.8

488.6

164

Shear

315 167

Moment -99.4

+

+ _

_

A

-507

B C

-444

84– 3.4 ≅ 80.5

j m

9

kN4.3112x2x2

2.755.80VA =+

=

area (abcd)

VA’ = 311.4 – 800 = – 488.6 kN

To find the centroid of a trapezoid on the horizontal axis;

2B1B2B1B2

3Lx

++

= x: distance from shorter dimension

m01.12.755.80

2.755.80x232)abcd(x =

++

=

MA = 311.4 x 1.01 = 315 kN.m

kN2.3198002x8x2

4.595.80VBA =−+

=

VBC = 319.2 – 600 = -280.8 kN

m2.44.595.80

4.595.80x238)aefc(x =

++

=

m.kN4.996x8002.4x2x8x2

4.595.80M B −=−⎥⎦⎤

⎢⎣⎡ +

=

kN3366008002x14x2

5.435.80VCB =−−+

=

VC = 336 – 500 = -164 kN

Check the end point; 05006008002x16x2

2.385.80=−−−

+ OK

B = 2

x B2 B1

L

10

m7.75.435.80

5.435.80x23

14)aghc(x =++

=

m.kN1676x60012x8007.7x2x14x2

5.435.80MC =−−⎥⎦⎤

⎢⎣⎡ +

=

Slope of V b/w x = 2 and x = 8 m (488.6+319.2)/6=134.6 kN/m

Equation of V b/w x = 2 and x = 8 m V(x) = -488.6 + 134.6 (x) for V(x)=0 x=3.6 m

Base pressure @ x = 2 + 3.6 = 5.6 m 65.6 kN/m2

Similarly @ x = 10.7 m V = 0

Base pressure @ x =10.7 m 52.2 kN/m2

maximum points of the moment diagram;

j and m are the points where shear forces are equal to zero (i.e. moment is max)

the distance between point b and point j

m9.26.655.80

6.655.80x23

)6.32()ajkc(x =+++

=

m.kN5076.3x8009.2x2x)6.32(x2

6.655.80M AB −=−⎥⎦⎤

⎢⎣⎡ +

+=

m73.5)amnc(x =

MBC = -444 kN.m

h f

d c

b B a 52.2 65.6

e g j

k

m

n

C A